-> ³T = 2: V = 12 
-> ³ 
-> ³w0 = 13.13 + .62 * T - 13.95 * V ^ .16 + .486 * T + V ^ .16 
-> ³ 
-> ³w1 = 13.13 + .62 * T + (.416 * T - 13.95) * V ^ .16 
-> ³ 
-> ³w2 = 13.13 + (.62 + .486) * T + (1 - 13.95) * V ^ .16 
-> ³ 
-> ³w3 = 13.13 + .62 * T + (.416 * T - 13.95) * V ^ .16 
-> ³ 
-> ³PRINT w0, w1, w2, w3 
 
-> D:\USER\QBASIC>QBASIC.EXE DW.BAS 
-> -3.930531     -5.152553     -3.930531     -5.152553 
-> ^^^^^^^^^                   ^^^^^^^^^ 
->  PP                          SN 
 
 
-> Sergey 
  
w0 is Pat's formula, containing a typo. The final "+" should have been 
a "*". 
  
w1 and w3 are identical. They are my version of the formula, without 
the typo, and with the terms rearranged so that the exponentiation is 
done only once. 
  
w2 is your version, which is mathematically equivalent to Pat's 
erroneous one - although I am mystified as to why you didn't simplify 
the expressions inside the brackets. 
  
So, yes, if you stick in some values for T and V, you'll get the same 
answer for w0 and w2. w1 and w3 will, of course, agree with each other, 
since they're identical, but they won't agree with the other two. 
  
Now try: 
  
w4 = 13.13 + 0.62 * T - 13.95 * V ^ 0.16 + 0.486 * T * V ^ 0.16 
  
This is what Pat *meant* to write, without the typo. When you put in 
some values for T and V, I bet you'll get the same as w1 and w3, which 
are my formula. 
  
                         dow 
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